Acid/Base stuff
Beauchamp
1
Problems – You should be able to match a pKa value with its acid in each group below and explain the differences.
You should be able to draw an arrow-pushing mechanism with general base, B:- for any of the acids, H-A. Include
resonance structures whenever appropriate. If there was a reaction shown between any two conjugate acids and
bases, you should be able to qualitatively and quantitatively indicate which side of the equilibrium is favored, and
what an approximate G is for the reaction. This is a big deal! You are learning the logic of organic chemistry,
right here.
acidic proton varies (on C, N, O or F)
H
1
H
C
N
H
H
pKa = 50
H
H
O
H pKa = 35
H
H
H
F
pKa = 16
Strongest acid has the lowest pKa.
The negative charge in the conjugate
base is on the most electronegative
atom, because they are all about
the same size.
pKa = 3
What is Keq and G for the following reactions?
H
C
H
H
equilibrium lies
completely to the left
H
N
H
weaker acid
H
H
H
weaker base
Ka1 = 10-50
pKa1 = 50
Ka1
Keq =
H
C
=
H
stronger base
stronger acid
Ka2 = 10-16
pKa2 = 16
-13
10-37
Ka2
H
H
-50
10
N
= 10
G = (pKa1 - pKa2) x 1.4 = (50 - 37) x 1.4 = (13) x 1.4 = +18 kcal/mole
H
H
N
O
stronger acid
equilibrium lies
completely to the right
H
Ka1 = 10
pKa1 = 16
Keq =
N
H
O
H
stronger base
-16
H
H
H
weaker acid
weaker base
Ka1
=
Ka2
10-16
10-37
Ka2 = 10-37
pKa2 = 37
= 10+21
G = (pKa1 - pKa2) x 1.4 = (16 - 37) x 1.4 = (-21) x 1.4 = -29 kcal/mole
2
acidic proton varies (on C, N, or O, in this problem F is not attached to a proton)
H2
H2
H2
H
H
C
H
C
C
H3 C
C
H3 C
N
H3 C
O
F
H3 C
H
pKa = 16
pKa = 50
pKa = 37
pKa = ?
H
H
H2
C
acidic proton is always on oxygen
3
I
O
H
Br
pKa = 11
O
Cl
H
O
F
O
H
not stable
pKa = NA
H
pKa = 7.5
pKa = 8.7
Strongest acid is ethanol, negative
charge is on the most electronegative
atom. Fluorine only makes one bond
and that is with carbon, not with a
proton, though the presence of the
F probably makes the C-H bonds more
acidic.
The strongest acid is hypochlorous acid.
All conjugate bases have the negative
charge on the oxygen, but the inductive
withdrawing effect of the chlorine is
greatest because it is the most electronegative.
acidic proton is always on oxygen
4
H
Cl
H
O
pKa's = 16
weakest
Cl
H
O
O
pKa's = 14.3
Cl
pKa's = 12.8
Cl
H
O
Cl
Cl
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pKa's = 12.2
strongest
More chlorines = more
inductive withdrawal,
which better helps stabilize
the anion conjugate base.
Acid/Base stuff
Beauchamp
acidic proton is always on carbon
H
Cl
5
H
H
C
H3C
O
C
H
H
Cl
H
pKa  32* H
C
H
Cl
O
H
pKa = -7
Br
pKa = 25 Cl
strongest
pKa = 16
H
weakest
S
H
H
pKa = 7
CH3
H3C
CH
C
H
H3C
O
I
H
H
H
pKa = 16
weakest
Se
H
H
pKa = 4
Te
H
pKa = 3
H
strongest
O
H
pKa = 10
All conjugate bases
have a minus on
oxygen. R groups are
inductively donating
which destabilizes the
conjugate base.
Fluorine is most electronegative, but
minus charge is more delocalized on
the larger iodide. Delocalization of the
minus charge is more important in this
example. All of the halogens have a Zeff = +7.
O
H
H
O
pKa's = 19
weakest
pKa = -10
strongest
pKa = -9
acidic proton is always on oxygen
O
* = my estimate
CH3
acidic proton varies (on O, S, Se or Te)
H
similar explanation to
previous problem.
H
pKa's = 17
pKa's = 15.8
pKa = 3
weakest
10
Cl
acidic proton varies (on F, Cl, Br or I)
F
9
C
Cl
acidic proton is always on oxygen
H2
H
H
C
H3C
H3C
O
pKa's = 15.5
strongest
8
Cl
pKa  40* H
pKa = 50 H
weakest
6
2
O
pKa = 4
strongest
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
similar explanation to
previous problem,
though all Zeff = +6.
All have minus on oxygen (good). The
first conjugate base has no resonance
structures, the second has four (one on
oxygen + three on carbon), the third has
two resonance structures with minus on
oxygen = best of this group.
Acid/Base stuff
Beauchamp
3
11 acidic proton is always on oxygen and another atom (C or N or a second O)
O
O
O
H
C
H3C
pKa = 20
weakest
H3C
H
12
H3C
N
pKa = 15
H
H
C
H
C
C
O
pKa = 5
strongest
H
acidic proton is always on two oxygen atoms
O
O
O
C
C
H
O
H2C
H
Ref. pKa = 4.7
I
H 2C
H
O
All three conjugate bases have minus on
oxygen (good), but the other resonance
structure is carbon (poorest), nitrogen
(middle) or another oxygen (best) for
stabilizing negative charge.
C
C
H
H 2C
pKa's = 2.59
pKa's = 3.2
H2C
O
H
O
pKa's = 2.85
C
H
H2C
Br
Cl
strongest
O
C
H
H2C
O
F
O
O
CH3
pKa's =2.90
O
pKa's = 4.8
weakest
Halogen atoms are electron withdrawing (F>Cl>Br>I) and inductively help stabilize minus in conjugate base. R groups are
inductively electron donating, which destabilizes minus charge (less stable = higher pKa).
13
acidic proton is always on two oxygen atoms
O
O
C
H3C
pKa = 5
weakest
H
Cl
O
C
C
H2
C
Cl
H
CH
O
O
O
C
Cl
H
O
C
O
Cl
pKa = 0.7
strongest
Cl
pKa = 2.8
Cl
pKa = 1.3
acidic proton is always on two oxygen atoms
O
O
O
Cl
H
H
H
O
O
O
Cl
pKa = 4.8
pKa = 4.5
pKa = 4.0
weakest
More chlorines = more
H inductive withdrawal,
which better helps stabilize
the anion conjugate base.
14
15
H
O
pKa = 2.8
Cl strongest
pKa's = 10
Only one chlorine atom in
each conjugate base. It has
the strongest effect when it
is closest to the negative
charge of the conjugate base.
H
strongest
acidic proton varies (on C, N, or O)
O
N
H
H
H
C
H
H
pKa's = 41
pKa's = 28
H
C
O
N
H
More electronegative atoms stabilizes
the negative charge best (O>N>C) and
more resonance delocalization stabilizes
the minus better than no delocalization.
H
H
H
pKa's = 18
16
O
pKa's = 37
H
acidic proton is always on two oxygen atoms
H3C
pKa's = 50
weakest
N
F
O2N
H
O
C
O
O
O
C
O
C
C
C
O
H
O
H
C
O
H
O
H
pKa = 4.3
pKa = 3.9
pKa = 3.5
pKa = 3.6
Ref. pKa = 4.2
weakest
strongest
All conjugate bases have negative charge spread over two oxygen atoms. Stronger electron withdrawal stabilizes minus better and donation destabilizes it.
O
H
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
Acid/Base stuff
Beauchamp
4
N
17
C
N
C
N
H
N
H
N
pKa = 23.6
strongest
minus delocalizes
onto nitrogen (good)
H
pKa = 28
weakest
H
pKa = 24.8
inductive withdrawal
by nitrile group
The acidic proton is always on
nitrogen and in all cases
delocalized into the ring. In the
first example it is also delocalized
(by resonance) onto the second
nitrogen (good), which is not
possible in the second example.
acidic proton is always on carbon and sometimes on one or two oxygen atoms
O
O
O
19
H H
pKa = 50
weakest
20
H
H
F3C
O
O
CF3
pKa = 9
H H
resonance on two
oxygen atoms
pKa = 20
H H
resonance on one
oxygen atom
pKa = 5
H H
resonance on two oxygen
atoms, plus inductive effect
of fluorine atoms
acidic proton is always on oxygen and has resonance in the ring
O
O
O
O
H
H
H
N
O
O
H
O2N
pKa's = 7.1
pKa's = 10.2
weakest
inductive donation
pKa's = 10.0
reference
pKa's = 8.4
inductive withdrawal
strongest
inductive withdrawal, plus extra
resonance with minus charge
on second oxygen of the NO2 group.
acidic proton is always on oxygen and has resonance in the ring
21
O
O
H Cl
H
O
O
H
H
Cl
22
C
H
H2 C
O
C
H
O
pKa's = 9.0
acidic proton is always on two oxygen atoms
O
O
O
O
Cl
pKa's = 10.0
weakest
reference
pKa's = 9.4
H3C
pKa = 2.85
strongest
C
H
C
H2 C
O
pKa = 4.7
reference
C
O
O
pKa's = 85
inductive withdrawal, closest
The neutral diacid is more electron
withdrawing (stronger acid than reference),
while the anion conjugate base is more
H
electron donating (weaker acid than
O
reference). The first negative charge
makes it harder to form the second
pKa = 5.70 negative charge.
weakest
acidic proton is always on carbon
23
pKa = 50
weakest
Only one chlorine
atom in each conjugate
base. It has the
strongest effect
when it is closest
to the negative
charge of the
conjugate base.
H
H
H
H
H
H
pKa = 42
two resonance structures
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pKa = 15*
* also aromatic
strongest
five resonance structures
Acid/Base stuff
24
Beauchamp
acidic proton is always on carbon
H
H
H
C
C
H
C
C
H
C
C
H
pKa = 25
H
H
strongest
H
H
pKa = 44
pKa = 50 weakest
H
25
5
H
H
acidic proton is always on nitrogen
H
C
N
H
H
C
N
H
H
H
pKa = -10
strongest
H
pKa = 5
The electronegativity of hybrid orbitals is: sp (50% s) >
sp2 (33% s) > sp3 (25% s). Greater electronegativity is better
able to stabilize the negative charge in the conjugate base, so
sp C-H bonds are the most acidic of these hydrocarbons, then
sp2 C-H and lastly sp3 C-H (lowest acidity of any acid in our
course). Remember pKa units are approximately exponents.
H
C
N
similar explanation to
previous problem.
H
H
H
pKa = 9 weakest
Bascity
Use full headed arrows to show two electron movement. Water is the reference acid in usual Kb tables. We won't
use pKb values. Instead, we will compare pKa values and judge bases to be stronger when their acids are weaker
(larger pKa) and bases to be weaker when their acids are stronger (lower pKa). In the examples that follow pair up
each base with the pKa value of its conjugate acid.
B
H
judge the strength of the base
from the weakness of the acid
O
B
H
use this acid's pKa value to judge
the electron donating power of B:
to the reference acid, H2O.
O
H
H
Here are a few qualitative examples. Where is the most basic site in each molecule below? Can an order of
basicity be explained for some or for all of the bases (approximate pKa’s of the conjugate acids are provided)?
O
O
C
C
H
H3C
CH3
H3C
O
pKa  -7
O
O
C
C
O
H3C
pKa  -7
H3 C
CH3
pKa  -6
pKa  -0.5 H
R
Use the lone pair of the C=X: as the most basic site.
O
H
N
N
N
C
H
N
H
N
C
H
N
N
O
H
H
pKa  0.2
pKa  5 (my guess)
H pK  7
a
H
N
N
H
H
pKa  13.6
The double bond lone pair is the most basic site. In the ketone it is the only lone pair. In the other bases there will be resonance
structures if you use the double bond lone pair and nitrogen is a better donor than oxygen and 2x nitrogen is better than 1x nitrogen.
R
H
N
R
H
H
C
H
N
N
N
H
H
H
H
R
N
N
C
H
R
A
H
C
H
H
N
H
C
H
H
N
N
N
N
N
H
H
H
H
H
carbon is positive too
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
Acid/Base stuff
Beauchamp
6
Problem – W at is the order of basicity among the following molecules of each group (1 = most basic)? Explain
your reasoning. Match the given pKa values with the conjugate acids of the indicated bases. Write arrow-pushing
mechanisms with general acid, H-A to illustrate the reactions. Include resonance structures whenever appropriate.
Where is the most basic site in each molecule? Explain your reasoning using arguments of inductive effects
(electronegativity), resonance effects (electron delocalization) or both. For any reaction between two conjugate
acids and bases, you should be able to qualitatively and quantitatively indicate which side of the equilibrium is
favored, and what is an approximate G is for the reaction.
1
H
H
electron pair donation varies (on C, N, O or F)
C
H
H
A
N
H
strongest
base
H
C
H
2
H2
C
H
C
H
O
H
F
A
H
N
H
O
H
H
F
H
pKa = 16
pKa = 3
electron pair donation varies (on C, N, O or F)
H2
H2
C
C
A HC
N
H
A
3
H3C
H
H
weakest
base
H
H
pKa = 35
H
pKa = 50
strongest
base
A
H
H
H3C
H
A
Strongest base has the
negative charge is on
the least electronegative
atom (smallest Zeff), so it
is best able to donate (share)
its electron density (because
they are all about the same
size). The most
electronegative atom (F) is
best able to carry the
negative charge and least
basic.
H2
C
H
O
A
H3C
H
F
A
pKa = NA
H
weakest
base
H2
C
H3C
H2
C
H
C
pKa = 50
H3C
H
H2
C
H
N
H3C
O
pKa = 16
pKa = 37
H
H2
C
H
H3C
H
F
H
pKa = ? (very negative)
Strongest base has the negative charge is on the least electronegative atom (C), so it is best able to donate (share) its electron density
(because all the atoms are about the same size). Fluorine is neutral in this example and not expected to be basic.
3
R
H
N
C
H3C
A
C
H
H3C
N
R
H
C
N
pKa = 7 H
H3C
pKa = 0
H
A
H
O
weakest base = 2x
oxygen donating pKa = -7
Remember, pKa is of
conjugate acid (with the
proton on).
H
N
H
N
H
C
H3C
R
H
C
H
O
H
N
N
H 3C
A
N
2nd best base = 1x H
nitrogen & 1x oxygen
donating
best base = 2x H
nitrogen donating
R
H
O
C
H3C
H
N
Remember, pKa is of conjugate acid (with the
proton on). The largest pKa is the weakest
acid, having the most basic conjugate base.
H
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
Acid/Base stuff
Beauchamp
7
4 H
C
H
N
H
A
H
H
N
A
H
N
H
pKa = 5
H
C
N
H
H
H
pKa = -10
H
5
N
H
H
strongest base
H
H
C
C
weakest base
H
H
C
A sp3 obitals are less electronegative
and better able to share electron
density (more basic) because they
are only 25% s. sp orbitals are
more electronegative and less able
to share electron density (50% s).
H
H
H
C
N
H
H
N
H
H
pKa = 9
R
N
H
H
A
H
N
A
N
N
N
H
H
best base = 3x
nitrogen donating
H
2nd best base = 2x
nitrogen donating
pKa = 13
A
Donating nitrogen atoms
are all sp2. but more
3rd best base = 1x
delocalization with full
nitrogen donating
octets is better for conjugate
acid (= weaker acid and
pKa = 5 stronger base)
H
C
pKa = 7
O
6
H
O
A
H
O
2nd best base
best base
pKa = 10
A
H
O
3rd base
pKa = 16
A
pKa = 5
The most stable anion (3rd structure) is the least basic and the least stable anion (middle structure) is sthe most basic. Stability
based on number of resonance structures and the electronegativity of the atoms (2x oxygen > 1x oxygen + 3x carbon > 1x oxygen).
7
O
O
O
O
O
F3C
CF3
H
H
pKa = 9
H
H
most basic pK = 50
a
least basic pKa = 5
pKa = 20
These are the conjugate bases of the acids in example 19 above. The most stable one (resonance + inductive withdrawal) is the
least basic. The least stable one is the most basic
6
O
O
C
H3C
O
H
A
O
C
H3C
C
H
N
H
A
least basic
pKa = 5
Less electronegative atoms
are less stable with negative
charge and more basic.
H3C
CH2 H
A
most basic
pKa = 15
pKa = 20
These are the conjugate bases of the acids in example 11 above. The most stable one (resonance = 2x oxygen) is the
least basic. The least stable one is the most basic
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
Acid/Base stuff
9
Beauchamp
H
H
H
8
H
C
C
H
A
C
C
H
A
C
H
least basic
H
H
most basic
H
C
H
H
pKa = 25
pKa = 50
A
pKa = 44
These are the conjugate bases of the acids in example 24 above. The most stable one is sp (most electronegative hybrid orbital) and
the least basic. sp3 orbital is least electronegative orbital and the worst for carrying the negative charge, thus most basic.
10
H
H
A
H
H
A
A
H
H
least basic
most basic
pKa = 15
pKa = 42
pKa = 50
These are the conjugate bases of the acids in example 23 above. The most stable one (resonance = 5x carbon) is the
least basic. The least stable one (1x carbon) is the most basic
Acid/Base arrow pushing worksheet
1. These proton transfer reactions are the first step in multistep mechanisms to be studied later in the course.
Supply the necessary curved arrows, lone pairs of electrons and/or formal charge to show how the first
step each reaction proceeds. Starting at e are simple proton transfer reactions generating a carbanions
(very important for organic chemistry). Generally, there is some stabilizing feature that allows a
carbanion to form via acid/base chemistry, such as inductive and/or resonance effects, but this is not
always the case. In working the problem below, show any important resonance structures or identify the
inductive effect that makes the reaction possible. Estimate a Keq for each reaction using the Ka’s.
a.
O
O
Na
H
sodium hydride
10-16
Keq =
= 10+21
-37
10
R
H
alcohols
Ka = 10-16
R
H-H
Na
Ka = 10-37
Equilibrium lies completely to the right.
b.
S
R
Na
O
H
sodium hydroxide
10-9
Keq =
= 10+7
-16
10
H
thiols
Ka = 10-9
H
S
R
O
Na
Equilibrium lies completely to the right.
H
Ka = 10-16
c.
Na
O
H
sodium
hydroxide
O
C
O
R
carboxylic acids
Ka = 10-5
H
Keq =
10-5
10-16
= 10
O
O
H
resonance
C
R
+11
C
O
R
O
O
Equilibrium lies completely to the right.
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
H
Ka = 10-16
Acid/Base stuff
Beauchamp
9
d.
e.
f.
O
R
Li
R
R
C
H
N
O
C
H
C
Keq =
10
R
O
C
ester
Ka = 10-25
H
LDA
-78 oC
O
O
C
resonance
H
H
R2NH
Ka = 10-37
H
-25
-37
R
O
C
H
Equilibrium lies completely to the right.
= 10+12
10
g.
O
R
N
H
Li
R
O
O
-78 oC
C
C
R
N
tertiary amide
H
Ka = 10-30
H
LDA
H
R
Keq =
10-30
-37
R
C
C
= 10+7
N
C
resonance
R
C
N
H
R
H
H
R
R2NH
Ka = 10-37
Equilibrium lies mostly to the right.
10
h.
H
R
N
R
H
Li
LDA
C
H
C
N
nitrile
Ka = 10-30
Keq =
-78 oC
H
C
C
N
resonance
H
10
H
C
H
-30
10
-37
= 10+7
Equilibrium lies mostly to the right.
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
C
N
R2NH
Ka = 10-37
Acid/Base stuff
Beauchamp
10
i.
O
O
C
C
H3C
H
C
C
Na
etc.
CH3
Ka = 10-37
H
10-9
Keq =
H-H
etc.
C
Ka = 10-9
H
sodium
hydride
O
H3 C
CH3
1,3-dicarbonyl
C
H
-78 oC
O
10-37
= 10+28
Equilibrium lies completely to the right.
j.
k.
Li
S
H2
C
H
S
CH3
H2C
C
H2
H
H
-78 oC
S
dithiane
Ka
10-35
10-35
10-50
CH2CH2CH2CH3
Ka = 10-50
S
n-butyl lithium
Keq =
H
Equilibrium lies completely to the right.
= 10+15
l.
H
Ph
Li
S
C
H2
C
H
H 2C
Ph
CH3
H
-78 oC
S
n-butyl lithium
sulfur ylid
-35
Ph = phenyl Ka  10
Keq =
10-35
10
-50
H
Ph
C
H2
C
possible
resonance
with sulfur
H
d orbitals?
S
H
C
H
H2
C
H2C
C
H2
Ph
Ph
= 10+15
H
Ph
CH3
Ka = 10-50
Equilibrium lies completely to the right.
m.
Br
Ph
Ph
P
Li
H
C
H
H2
C
CH3
H2C
Ph
H
phosphorous ylid
(Ph = phenyl) Ka  10-35
C
H2
n-butyl lithium
H
Ph
-78 oC
Ph
P
C
possible
Ph
resonance
H with phosphorous Ph
P
d orbitals?
Ph
Keq =
10-35
10-50
= 10
+15
Equilibrium lies completely to the right.
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
Ph
H
H
C
H
H2
C
H2C
C
H2
Ka = 10-50
CH3
Acid/Base stuff
Beauchamp
11
2. Lone pair donors to very strong acid
All of the following functional groups react with strong protic acid as the first step of a reaction studied in organic
chemistry. Often subsequent chemistry occurs after that initial step and you will study most of those reactions later
in the course. Show how they react in the first step by including all lone pairs, curved arrows to show electron
movement and formal charge.
Acid/Base arrow pushing worksheet
lone pair donors
lone pair acceptor
O
O
H
Keq = ?
H
O
H
H
O
S
OH
H
N
H
H
H
ammonia
O
103
2
O
H
S
= 10+1
OH
equilibrium
O
Equilibrium lies to the right.
10
H
102
10-9
H
Keq = ?
pKa's = -2, 9
H
Keq =
O
pKa's = -3, -2
O
Keq =
H
H
N
H
H
O
H
equilibrium
H
= 1011
Equilibrium lies completely to the right.
H
R
N
H
H
O
H
Keq = ?
R
H
H
1o amine
Keq =
N
H
H
O
H
equilibrium
H
equilibrium
H
equilibrium
pKa's = -2, 10
102
10-10
= 1012
H
Equilibrium lies completely to the right.
H
R
N
H
H
O
H
Keq = ?
R
H
R
2 amine
o
Keq =
N
H
H
O
pKa's = -2, 10
102
10-10
R
= 1012
Equilibrium lies completely to the right.
H
R
N
R
H
O
H
Keq = ?
R
R
3o amine
H
Keq =
N
R
H
O
pKa's = -2, 10
102
10-10
R
= 1012
Equilibrium lies completely to the right.
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
Acid/Base stuff
Beauchamp
H
R
O
O
additional
chemistry
possible
H
pKa's = -2, -3
H
R
O
H
H
O
H
2
10
Keq =
H
O
Keq = ?
H
H
alcohol
R
12
O
= 10-1
3
10
Equilibrium is pretty even
Keq = ?
H
additional
chemistry
possible
H
R
pKa's = -2, -3
H
ether
10
Keq =
O
H
= 10-1
103
H
epoxide (ether)
O
R
H
H
H
102
= 10-1
103
H
O
H
H
R
C
R
C
Keq = ?
H
H
R'
additional
chemistry
possible
O
resonance
(2)
O
O
H
Equilibrium is pretty even
O
H
C
R'
R
R'
pKa's = -2, -7
ketone
Keq =
102
106
= 10-4
Equilibrium lies a little to the left
H
R
C
N
H
O
additional
chemistry
possible
O
pKa's = -2, -3
Keq =
O
Equilibrium is pretty even
Keq = ?
H
O
R
2
SO3H
nitrile
Keq = ?
N
C
pKa's = -3, -10
R
sulfuric acid
Keq =
103
1010
= 10-7
H
resonance
(2)
N
C
R
Equilibrium lies to the left but still generates some
of the protonated nitrile that can react further.
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
additional
chemistry
possible
Acid/Base stuff
Beauchamp
13
H
O
H
C
O
H
H
R
O
carboxylic acid
Keq =
10
H
R
C
O
R
C
= 10-5
107
O
Keq = ?
O
pKa's = -2, -6
C
H
H
ester
10
Keq =
R
2
= 10
106
-4
H
H
NH2
Keq = ?
pKa's = -2, 0
amide
Keq =
10
R'
O
O
C
C
R
R'
R
O
O
H
H
O
O
O
C
C
C
NH2
H
N
H
C
O
H
NH2
Keq =
R
102
10-12
NH2
R
H
H
NH2
additional
chemistry
possible
= 10+2 Equilibrium lies a little to the right
H
R
R
2
100
additional
chemistry
possible
H
O
C
R
R'
O
H
H
O
additional
chemistry
possible
Equilibrium lies a little to the left
O
H
H
Equilibrium lies a little to the left
R'
R
R
O
H
O
C
H
2
O
H
O
O
C
pKa's = -2, -6
H
H
H
O
Keq = ?
H
Keq = ?
N
pKa's = -2, 12
C
H
= 10
+14
R
NH2
H
H
N
N
C
C
R
NH2
Equilibrium lies completely to the right
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC
R
NH2
additional
chemistry
possible
Acid/Base stuff
Beauchamp
14
Carbon-carbon pi bonds as weak electron pair donors to very strong acid
H
H
C
R
alkene
SO3H
O
H
Keq = ?
C
pKa's = -3, -10
CH2
sulfuric acid
103
Keq =
1010
R
 10-7
H
C
H2
Equilibrium lies to the left but still generates some
of the protonated alkene that can react further.
R
H
O
H
Keq = ?
C
H
CH2
enol ether
pKa's = -2, -7
102
Keq =
C
C
H
H
O
R
10
 10-7
H
Keq = ?
C
pKa's = -10, -10
C
C
H
aromatic
H
E+ = electrophile
(Lewis acid = electron
pair acceptor)
10
Keq =
10
1010
1
C
H2
H
R
C
H
C
C
H
C
additional
chemistry
possible
H
H
C
H
E
C
R
H
E = electrophile
H
additional
chemistry
possible
H
Equilibrium lies to the left but still generates some
of the protonated alkyne that can react further.
10
H
C
C
C
H2
pKa's = -3, -10
103
Keq =
C
resonance
(2)
Keq = ?
SO3H
sulfuric acid
C
H
 10-5 Equilibrium lies a little to the left
107
alkyne
H
O
O
C
R
R
R
O
R
additional
chemistry
possible
carbocation
chemistry
H
E
C
C
H
H
H
H
H
C
C
C
C
C
C
H
H
C
H
E
H
H
C
C
C
C
C
E additional
chemistry
possible
H
H
Equilibrium is even (not really sure about pKa of electrophile, but they are very reactive)
At this point we are mainly interested in understanding acid/base proton transfers, curved arrow
pushing, formal charge, recognizing resonance structures and using the logic arguments of inductive
effects and resonance effects to explain relative stabilities of acids and bases. If you can do these things,
you are well on your way to understanding organic chemistry and biochemistry.
Tautomers - There are many worked examples in the tautomer section. Use those to practice with and then try
some possibilities without worked answers. Ask me if you get stuck.
y:\files\classes\315\315 Handouts\315 Fall 2013\1b acid-base list, 315 answers, 14p.DOC